Question

The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and square that produce a minimum area.

I would like to solve it by incorporating A=(sqrt(3)x^2)/4 (area of an equilateral triangle)

Answer 1

let the sides of the triangle be x and sides of the square be y

3x+4y=10

total area of the 2 shapes is = (sqrt(3)x^2)/4 + y^2

4y=10-3x
y=10-3x/4

.`. A = (sqrt(3)x^2)/4 + (10-3x/4)^2
= (sqrt(3)x^2)/4 + (100 - 60x + 9x^2)/16
dA/dx = sqrt3x/2 - 60/16 + 18x/16

let dA/dx=0 to find critical point

sqrt(3)x/2+18x/16=60/16

8sqrt3x+18x=60
x=60/(8sqrt3+18)

that x value will give the minimum area and sub it back into 3x+4y=10 to get y

Answer 2

How did you find dA/dx?